Xor-Grid Problem
Description
Given a matrix $a$ of size $n \times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.
Let's define $f(i)$ and $g(j)$ as the XOR of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:
- Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \le j \le m$; or
- Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \le i \le n$.
An example of applying an operation on column $2$ of the matrix. In this example, as we apply an operation on column $2$, all elements in this column are changed:
- $a_{1,2} := f(1) = a_{1,1} \oplus a_{1,2} \oplus a_{1,3} \oplus a_{1,4} = 1 \oplus 1 \oplus 1 \oplus 1 = 0$
- $a_{2,2} := f(2) = a_{2,1} \oplus a_{2,2} \oplus a_{2,3} \oplus a_{2,4} = 2 \oplus 3 \oplus 5 \oplus 7 = 3$
- $a_{3,2} := f(3) = a_{3,1} \oplus a_{3,2} \oplus a_{3,3} \oplus a_{3,4} = 2 \oplus 0 \oplus 3 \oplus 0 = 1$
- $a_{4,2} := f(4) = a_{4,1} \oplus a_{4,2} \oplus a_{4,3} \oplus a_{4,4} = 10 \oplus 11 \oplus 12 \oplus 16 = 29$
You can apply the operations any number of times. Then, we calculate the $\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.
More formally, $\textit{beauty}(a) = \sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.
Find the minimum $\textit{beauty}$ among all obtainable matrices.
The first line contains a single integer $t$ ($1 \le t \le 250$) — the number of test cases.
The first line of each test case contains two integers $n$ and $m$ ($1 \le n, m \le 15$) — the number of rows and columns of $a$, respectively.
The next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \ldots, a_{i,m}$ ($0 \le a_{i,j} < 2^{20}$) — description of the matrix $a$.
It is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.
For each test case, print a single integer $b$ — the smallest possible $\textit{beauty}$ of the matrix.
Input
The first line contains a single integer $t$ ($1 \le t \le 250$) — the number of test cases.
The first line of each test case contains two integers $n$ and $m$ ($1 \le n, m \le 15$) — the number of rows and columns of $a$, respectively.
The next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \ldots, a_{i,m}$ ($0 \le a_{i,j} < 2^{20}$) — description of the matrix $a$.
It is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.
Output
For each test case, print a single integer $b$ — the smallest possible $\textit{beauty}$ of the matrix.
4
1 2
1 3
2 3
0 1 0
5 4 4
2 3
0 2 4
4 5 1
3 3
1 2 3
4 5 6
7 8 9
1
3
13
24
Note
Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.
In the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \oplus 3 = 2$. Then, we'll receive this matrix:
| 2 | 3 |
In the second test case, you can apply an operation $r(1)$, which assigns:
- $a_{1,1} := g(1) = 0 \oplus 5 = 5$
- $a_{1,2} := g(2) = 1 \oplus 4 = 5$
- $a_{1,3} := g(3) = 0 \oplus 4 = 4$
The resulting matrix after performing the operation is:
| 5 | 5 | 4 |
| 5 | 4 | 4 |
In the third test case, the best way to achieve minimum $\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:
| 0 | 4 | 6 |
| 4 | 5 | 6 |